document.write( "Question 773179: what is the length and width if the area of a rectangle is 54 ft squared and the perimeter is 42ft \n" ); document.write( "

Algebra.Com's Answer #471457 by Cromlix(4381) $\"\"$ $\"About$

You can put this solution on YOUR website!
Length * Width = Area = 54ft
\n" ); document.write( "L * W = 54 => W = 54/L
\n" ); document.write( "Perimeter = 2* Length + 2* Width
\n" ); document.write( "Perimeter = 2L + 2W
\n" ); document.write( "Replace W with 54/L
\n" ); document.write( "Perimeter = 2L + 2(54/L)
\n" ); document.write( "42ft = 2L + 108/L
\n" ); document.write( "Common denominator = L
\n" ); document.write( "42L = 2L^2 + 108
\n" ); document.write( "Rearrange.
\n" ); document.write( "2L^2 - 42L + 108 = 0
\n" ); document.write( "Divide by 2
\n" ); document.write( "L^2 - 21L + 54 = 0
\n" ); document.write( "(L - 3)(L - 18) = 0
\n" ); document.write( "L - 3 = 0
\n" ); document.write( "L = 3
\n" ); document.write( "L - 18 = 0
\n" ); document.write( "L = 18
\n" ); document.write( "therefore either the length
\n" ); document.write( "= 18ft and the width = 3ft.
\n" ); document.write( "OR
\n" ); document.write( "the length = 3ft and the width
\n" ); document.write( "= 18ft.
\n" ); document.write( "Hope this helps.
\n" ); document.write( ":-) \n" ); document.write( "

\n" );